Re: [Haskell-cafe] Convert IO Int to Int

On 2009/06/09, at 19:33, Tobias Olausson wrote:

You can not convert an IO Int to Int, or at least, you shouldn't. However, you can do as follows:

test :: IO () test = do int <- randomRIO -- or whatever it is called print $ useInt int

useInt :: Int -> Int useInt x = x+10

Or, you can lift pure function into IO. the below test' function almost same as above test function. (But I used randomIO instead of randomRIO because it seemed to be a typo :-) test' = print =<< fmap useInt randomIO I think it is more handy than using do notation, when you want to do something simple with monads. And converting IO Int to IO anything is much easier and safer than converting IO Int to Int. ghci> :m +System.Random Data.Char ghci> :t fmap (+1) randomIO fmap (+1) randomIO :: (Num a, Random a) => IO a ghci> :t fmap show randomIO fmap show randomIO :: IO String ghci> :t fmap chr randomIO fmap Data.Char.chr randomIO :: IO Char ghci> :t fmap (+) randomIO fmap (+) randomIO :: (Num a, Random a) => IO (a -> a) Thanks, Hashimoto

//Tobias

2009/6/9 ptrash :

...

Hi,

I am using the System.Random method randomRIO. How can I convert its output to an Int?

Thanks... -- View this message in context: http://www.nabble.com/Convert-IO-Int- to-Int-tp23940249p23940249.html Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.

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